Question

integral (x^x)^x(2xlogx+x)

Answer 1

To find:

$\int (x^{x})^{x}(2xlogx+x)dx$

Step-by-step explanation:

Let, $(x^{x})^{x}=z$

$\Rightarrow x^{x^{2}}=z$

Taking $log$ in both sides, we get

$\quad log(x^{x^{2}})=logz$

$\Rightarrow x^{2}logx=logz$

Taking differentials, we get

$\quad d(x^{2}logx)=d(logz)$

$\Rightarrow (2xlogx+x)dx=\frac{dz}{z}$

Now, $\int (x^{x})^{x}(2xlogx+x)dx$

$=\int z\frac{dz}{z}$

$=\int dz$

$=z+c$ where $c=$ integral constant

$=(x^{x})^{x}+c$

This is the required integral.

Answer:

$\int (x^{x})^{x}(2xlogx+x)dx=(x^{x})^{x}+c$ where $c$ is integral constant.